为什么numpy cov对角元素和var函数具有不同的值? [英] Why do numpy cov diagonal elements and var functions have different values?

问题描述

In [127]: x = np.arange(2)

In [128]: np.cov(x,x)
Out[128]:
array([[ 0.5,  0.5],
       [ 0.5,  0.5]])

In [129]: x.var()
Out[129]: 0.25

为什么这是行为?我相信协方差矩阵对角线元素应该是级数的方差.

Why is this the behavior? I believe that covariance matrix diagonal elements should be the variance of the series.

推荐答案

在numpy中，cov的默认"delta自由度"为1，而var的默认ddof为0. c2>

In numpy, cov defaults to a "delta degree of freedom" of 1 while var defaults to a ddof of 0. From the notes to numpy.var

Notes
-----
The variance is the average of the squared deviations from the mean,
i.e.,  ``var = mean(abs(x - x.mean())**2)``.

The mean is normally calculated as ``x.sum() / N``, where ``N = len(x)``.
If, however, `ddof` is specified, the divisor ``N - ddof`` is used
instead.  In standard statistical practice, ``ddof=1`` provides an
unbiased estimator of the variance of a hypothetical infinite population.
``ddof=0`` provides a maximum likelihood estimate of the variance for
normally distributed variables.

因此，您可以通过以下方式使他们同意:

So you can get them to agree by taking:

In [69]: cov(x,x)#defaulting to ddof=1
Out[69]: 
array([[ 0.5,  0.5],
       [ 0.5,  0.5]])

In [70]: x.var(ddof=1)
Out[70]: 0.5

In [71]: cov(x,x,ddof=0)
Out[71]: 
array([[ 0.25,  0.25],
       [ 0.25,  0.25]])

In [72]: x.var()#defaulting to ddof=0
Out[72]: 0.25

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